Question

A moving coil galvanometer has 100 turns and each turn has an area of $2.0 \, \text{cm}^2$. The magnetic field produced by the magnet is $0.01 \, \text{T}$ and the deflection in the coil is $0.05$ radians when a current of $10 \, \text{mA}$ is passed through it. The torsional constant of the suspension wire is $x \times 10^{-5} \, \text{N} \cdot \text{m/rad}$. The value of $x$ is:

Answer 1

Correct Answer: 4

To determine the value of \(x\) for the torsional constant of the suspension wire, we begin by considering the dynamics of a moving coil galvanometer. The torque \(\tau\) on the coil due to current \(I\) is given by:
\[\tau = n \cdot B \cdot A \cdot I\]
where \(n\) is the number of turns, \(B\) is the magnetic field, \(A\) is the area of each turn, and \(I\) is the current.
Substituting the given values: \(n = 100\), \(B = 0.01 \, \text{T}\), \(A = 2.0 \times 10^{-4} \, \text{m}^2\) (converted from cm² to m²), and \(I = 10 \times 10^{-3} \, \text{A}\), we calculate:
\[\tau = 100 \times 0.01 \times 2.0 \times 10^{-4} \times 10 \times 10^{-3} = 2 \times 10^{-5} \, \text{N} \cdot \text{m}\]
The torsional constant \(k\) relates the torque to the angular deflection \(\theta\) as \(\tau = k \cdot \theta\). Given \(\theta = 0.05\) radians, we substitute to find \(k\):
\[2 \times 10^{-5} = k \times 0.05\]
Solving for \(k\):
\[k = \frac{2 \times 10^{-5}}{0.05} = 4 \times 10^{-4} \, \text{N} \cdot \text{m/rad}\]
The torsional constant is expressed as \(x \times 10^{-5}\), so equating \(4 \times 10^{-4} \) to \(x \times 10^{-5}\) gives:
\[x \times 10^{-5} = 4 \times 10^{-4}\]
\(x = 4\)

Answer 2

Given: - Number of turns: \(N = 100\) - Area of each turn: \(A = 2.0 \, \text{cm}^2 = 2.0 \times 10^{-4} \, \text{m}^2\) - Magnetic field: \(B = 0.01 \, \text{T}\) - Deflection: \(\theta = 0.05 \, \text{radian}\) - Current: \(I = 10 \, \text{mA} = 10 \times 10^{-3} \, \text{A}\)

Step 1: Calculating the Torque Acting on the Coil

The torque (\(\tau\)) acting on a moving coil galvanometer is given by:

\[ \tau = N \times B \times I \times A \]

Substituting the given values:

\[ \tau = 100 \times 0.01 \, \text{T} \times 10 \times 10^{-3} \, \text{A} \times 2.0 \times 10^{-4} \, \text{m}^2 \] \[ \tau = 100 \times 0.01 \times 0.01 \times 2.0 \times 10^{-4} \, \text{N} \times \text{m} \] \[ \tau = 2 \times 10^{-5} \, \text{N} \times \text{m} \]

Step 2: Relating Torque to the Torsional Constant

The torque is also related to the torsional constant (\(k\)) and deflection (\(\theta\)) by:

\[ \tau = k \times \theta \]

Rearranging to find \(k\):

\[ k = \frac{\tau}{\theta} \]

Substituting the given values:

\[ k = \frac{2 \times 10^{-5}}{0.05} \, \text{N} \times \text{m/rad} \] \[ k = 4 \times 10^{-4} \, \text{N} \times \text{m/rad} \]

Step 3: Finding the Value of \(x\)

Given that \(k = x \times 10^{-5} \, \text{N} \times \text{m/rad}\):

\[ 4 \times 10^{-4} = x \times 10^{-5} \]

Solving for \(x\):

\[ x = 4 \]

Conclusion: The value of \(x\) is 4.